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A congruence modulo p^2 for the p-th coefficient of e(1-z)^((1-z)/z)

Primarily AI-generated textHuman understanding: no partsmath.NT — Number Theory

Contributed by David Niedbala Giraudin ↗

Version 1 / Nov 04, 2026 / CC BY 4.0

Abstract

Define E_n by sum_{n>=0} E_n z^n = exp(sum_{k>=1} z^k/(k(k+1))) = e(1-z)^((1-z)/z). Equivalently, e(1+1/x)^(-x) = sum_{n>=0} E_n (x+1)^(-n), the reciprocal of the expansion of e^(-1)(1+1/x)^x used by Yang in refinements of Carleman's inequality. We prove that for every odd prime p, E_p = -p/2 (mod p^2); in particular v_p(E_p) = 1. The proof is elementary. With Phi(u) = e(1+u)^(-1/u), it combines the two expressions of E_p through the coefficients of Phi that come from the identities e(1-z)^((1-z)/z) = Phi(-z)/(1-z) and e(1-z)^((1-z)/z) = Phi(z/(1-z)), and it reduces the congruence to the differential recurrence of the E_n read modulo p. A script checks the statement and the intermediate congruences for all odd primes below 200 and a few larger ones.

Provenance statement

The research and the writing were carried out by the author with Claude (Anthropic). The proof was checked by an independent AI review and by exact computation. No professional mathematician has reviewed the note yet.

Tools used

Anthropic
Claude Opus

References

  1. Y. Hu, C. Mortici, On the coefficients of an expansion of (1+1/x)x(1+1/x)^x related to Carleman's inequality, arXiv:1401.2236 (2014).arXiv
  2. D. Niedbala Giraudin, The pp-adic valuations of the coefficients in Yang's expansion of (1+1/x)x(1+1/x)^x, preprint hexagon:2611.00074, https://hexagonmath.org (2026).hexagon
  3. The On-Line Encyclopedia of Integer Sequences, sequences A055505 and A055535, https://oeis.org.link
  4. X. Yang, On Carleman's inequality, J. Math. Anal. Appl. 253 (2001), 691--694.
  5. X. Yang, Approximations for constant ee and their applications, J. Math. Anal. Appl. 262 (2001), 651--659.

Version history

  1. v1 · 4 pagesSubmitted by David Niedbala GiraudinInitial depositCurrentNov 03, 2026